In many books, like Visual Complex Analysis. talk about the real original of complex number. the author begin with this equation:
$$x^3=15x+4$$
Then the author use the formula $$x=\sqrt[3]{q+\sqrt{q^2-p^3}}+\sqrt[3]{q-\sqrt{q^2-p^3}}$$
to say that the equation has a root $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$
Apparently, $x=4$ is a root of the equation $x^3=15x+4$. Then the author guess $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i} =4.$$ then introduction complex arithmetic. It seems very natural. but equation $x^3=15x+4$ has three different roots. why the author, or probably Bombelli guess this equal $4$, not other roots?
Best Answer
He (Bombelli) probably used the Cardano formula to obtain the $\textbf{one}$ root solution you just described $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$
He then had to find the cube roots of each of the radicals above. I am not sure exactly how he found them but he may have used the following sometimes used polynomial to find each of the cube roots
$$\frac{-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3}{-64} = 0$$ where $x$ = $2$ and $y$ = $11i$ to get
$$a^9 + \frac{3}{2}a^6 - \frac{3327}{64}a^3 + \frac{1}{8} = 0$$
This polynomial has one rational root $a$ = $2$
He could now take the rational root $a$ = $2$ he found above to find $b$ in the equation below in order to denest the two cube roots in Cardano's formula. The $2$ on the $RHS$ is the $2$ under the radical $$ a^3+3ab^2=2$$
$$2^3 + 3(2)b^2 = 2$$ $$b=\pm \sqrt\frac{-6}{6} = \pm i$$
So using the $a$ he found and the $b$ he found he could derive the cube roots as $a + b$. This gives the two results $$2 + i$$ and $$2 -i$$ Each is the cube root of the nested radicals in the Cardano solution.
So using the $a$ he found and the $b$ he found he could observe that this was equal to the real root of $4$ found using other methods
$$\left(2 + i\right) + \left(2 - i\right) = 4$$